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MCQs

Total Questions : 70 | Page 4 of 7 pages
Question 31. Find the heat flow rate through the composite wall as shown in figure. Assume one dimensional flow and take
k 1 = 150 W/m degree
k 2 = 30 W/m degree
k 3 = 65 W/m degree
k 4 = 50 W/m degree
AB = 3 cm, BC = 8 cm and CD = 5 cm.
The distance between middle horizontal line from the top is 3 cm and from the bottom is 7 cmFind The Heat Flow Rate Through The Composite Wall As Shown ...
  1.    1173.88 W
  2.    1273.88 W
  3.    1373.88 W
  4.    1473.88 W
 Discuss Question
Answer: Option B. -> 1273.88 W
Answer: (b).1273.88 W
Question 32. The oven of an electric store, of total outside surface area 2.9 m² dissipates electric energy at the rate of 600 W. The surrounding room air is at 20 degree Celsius and the surface coefficient of heat transfer between the room air and the surface of the oven is estimated to be 11.35 W/m² degree. Determine the average steady state temperature of the outside surface of the storeThe Oven Of An Electric Store, Of Total Outside Surface Area...
  1.    38.22 degree Celsius
  2.    48.22 degree Celsius
  3.    58.22 degree Celsius
  4.    68.22 degree Celsius
 Discuss Question
Answer: Option A. -> 38.22 degree Celsius
Answer: (a).38.22 degree Celsius
Question 33. A steel pipe of 20 mm inner diameter and 2 mm thickness is covered with 20 mm thick of fiber glass insulation (k = 0.05 W/m degree). If the inside and outside convective coefficients are 10 W/m² degree and 5 W/m² degree, calculate the overall heat transfer coefficient based on inside diameter of pipe. In the diagram, the diameter of small circle is 20 mmA Steel Pipe Of 20 Mm Inner Diameter And 2 Mm Thickness Is C...
  1.    1.789 W/m² degree
  2.    2.789 W/m² degree
  3.    3.789 W/m² degree
  4.    4.789 W/m² degree
 Discuss Question
Answer: Option B. -> 2.789 W/m² degree
Answer: (b).2.789 W/m² degree
Question 34. The accompanying sketch shows the schematic arrangement for measuring the thermal conductivity by the guarded hot plate method. Two similar 1 cm thick specimens receive heat from a 6.5 cm by 6.5 cm guard heater. When the power dissipation by the wattmeter was 15 W, the thermocouples inserted at the hot and cold surfaces indicated temperatures as 325 K and 300 K. What is the thermal conductivity of the test specimen material?The Accompanying Sketch Shows The Schematic Arrangement For ...
  1.    0.81 W/m K
  2.    0.71 W/m k
  3.    0.61 W/m K
  4.    0.51 W/m K
 Discuss Question
Answer: Option B. -> 0.71 W/m k
Answer: (b).0.71 W/m k
Question 35. In Cartesian coordinates the heat conduction equation is given byIn Cartesian Coordinates The Heat Conduction Equation Is Giv...
  1.    d²t/dx² + d²t/dy2 + d²t/dz2 + q g = (1/α) (d t/d T)
  2.    2d²t/dx² + d²t/dy2 + d²t/dz2 + 34q g = (d t/d T)
  3.    d²t/dx² + 3d²t/dy2 + d²t/dz2 = (1/α) (d t/d T)
  4.    4d²t/dx² + d²t/dy2 + d²t/dz2 + 1/2q g = (1/α) (d t/d T)
 Discuss Question
Answer: Option A. -> d²t/dx² + d²t/dy2 + d²t/dz2 + q g = (1/α) (d t/d T)
Answer: (a).d²t/dx² + d²t/dy2 + d²t/dz2 + q g = (1/α) (d t/d T)
Question 36. For the same amount of fabrication material and same inside capacity, the heat loss is lowest in
  1.    Sphere
  2.    Cylinder
  3.    Rectangle
  4.    Cube
 Discuss Question
Answer: Option B. -> Cylinder
Answer: (b).Cylinder
Question 37. A plane wall of thickness δ has its surfaces maintained at temperatures T₁ and T₂. The wall is made of a material whose thermal conductivity varies with temperature according to the relation k = k₀ T². Find the expression to work out the steady state heat conduction through the wall?
  1.    Q = 2A k₀ (T₁³ – T₂³)/3 δ
  2.    Q = A k₀ (T₁³ – T₂³)/3 δ
  3.    Q = A k₀ (T₁² – T₂²)/3 δ
  4.    Q = A k₀ (T₁ – T₂)/3 δ
 Discuss Question
Answer: Option B. -> Q = A k₀ (T₁³ – T₂³)/3 δ
Answer: (b).Q = A k₀ (T₁³ – T₂³)/3 δ
Question 38. With variable thermal conductivity, Fourier law of heat conduction through a plane wall can be expressed as
  1.    Q = -k0 (1 + β t) A d t/d x
  2.    Q = k0 (1 + β t) A d t/d x
  3.    Q = – (1 + β t) A d t/d x
  4.    Q = (1 + β t) A d t/d x
 Discuss Question
Answer: Option A. -> Q = -k0 (1 + β t) A d t/d x
Answer: (a).Q = -k0 (1 + β t) A d t/d x
Question 39. The inner and outer surfaces of a furnace wall, 25 cm thick, are at 300 degree Celsius and 30 degree Celsius. Here thermal conductivity is given by the relation
K = (1.45 + 0.5 * 10¯⁵ t²) KJ/m hr deg
Where, t is the temperature in degree centigrade. Calculate the heat loss per square meter of the wall surface area?
  1.    1355.3 kJ/m² hr
  2.    2345.8 kJ/m² hr
  3.    1745.8 kJ/m² hr
  4.    7895.9 kJ/m² hr
 Discuss Question
Answer: Option C. -> 1745.8 kJ/m² hr
Answer: (c).1745.8 kJ/m² hr
Question 40. The mean thermal conductivity evaluated at the arithmetic mean temperature is represented by
  1.    km = k0 [1 + β (t1 – t2)/2].
  2.    km = k0 [1 + (t1 + t2)/2].
  3.    km = k0 [1 + β (t1 + t2)/3].
  4.    km = k0 [1 + β (t1 + t2)/2].
 Discuss Question
Answer: Option D. -> km = k0 [1 + β (t1 + t2)/2].
Answer: (d).km = k0 [1 + β (t1 + t2)/2].

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